If 200 grams of water is to be heated from 24.0°C to 100.0°C to make a cup of tea, how much heat must be added? The specific heat of water is 4.18 J/g∙C

H = 63536J

Explanation:

mass = 200g

Initial Temperature = 24°C

Final Temperature = 100 °C

ΔT = Final - Initial = 100 - 24 = 76°C

Specific heat, c = 4.18 J/g∙C

Heat, H = ?

All these parameters are related by the equation;

H = mcΔT

Inserting the values;

H = 200 * 4.18 * 76

H = 63536J